AQA A-Level Biology Paper 1 7402-1 Question Paper June 2021 / AQA A-Level Biology Paper 1 7402/1 Question Paper June 2021 / AQA A-Level Biology Paper 1
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IB/H/Jun21/E14 7402/1
For Examiner’s Use
Question
Mark
1
2
3
4
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8
9
10
TOTAL
Time allowed: 2 hours
Materials For this paper you must have:
•a ruler with millimetre measurements
•a scientific calculator.
Instructions
•Use black ink or black ball-point pen.
•Fill in the boxes at the top of this page.
•Answer all questions.
•You must answer the questions in the spaces provided. Do not write
outside the box around each page or on blank pages.
•If you need extra space for your answer(s), use the lined pages at the end ofthis book. Write the question number against your answer(s).
•Show all your working.
•Do all rough work in this book. Cross through any work you do not want
to be marked.
Information
•The marks for the questions are shown in brackets.
•The maximum mark for this paper is 91.
Please write clearly in block capitals.
Centre number
Candidate number
Surname
Forename(s)
Candidate signature
I declare this is my own work.
A-level
BIOLOGY
Paper 1 AQA
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Answer all questions in the spaces provided.
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Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst.
[3 marks]
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Scientists investigated the action of the enzyme ATP synthase. They made reaction mixtures each containing:
•ATP synthase
•buffer (to control pH)
•substrates.
One of the substrates required in these reaction mixtures is inorganic phosphate (Pi).
Tick () one box to show which other substrate the scientists must add to the reaction mixtures to produce ATP.
[1 mark]
Adenine
Adenosine diphosphate
Glucose
Ribose
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The scientists investigated the effect of concentration of inorganic phosphate (Pi) on ATP synthase activity.
After 2 minutes, they stopped each reaction and then measured the concentration of ATP.
Figure 1 shows the scientists’ results.
Figure 1
Suggest and explain a procedure the scientists could have used to stop each reaction.
[2 marks]
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Explain the change in ATP concentration with increasing inorganic phosphate concentration.
[2 marks]
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Explain the advantage for larger animals of having a specialised system that
facilitates oxygen uptake.
[2 marks]
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Figure 2 shows two models of oxygen uptake found in animals.
Figure 2
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Suggest how the environmental conditions have resulted in adaptations of systems using Model A rather than Model B.
[2 marks]
Question 2 continues on the next page
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Figure 3 shows changes in concentration of oxygen in two gas exchange systems.
Figure 3
A student studied Figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system.
Use Figure 3 to justify this conclusion.
[2 marks]
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Explain how the counter-current principle allows efficient oxygen uptake in the
fish gas exchange system.
[2 marks]
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Table 1 shows features of two mammals.
Bats are flying mammals; shrews are ground-living mammals.
Table 1
Mammal
Mean body mass / kg
Mean lung volume / cm3
Bat
0.096
12.48
Shrew
0.024
0.72
Calculate how many times the lung volume per unit of body mass of the bat is greater than that of the shrew.
Give your answer to an appropriate number of significant figures.
Give one suggestion to explain this difference.
[3 marks]
Answer
Explanation
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Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation.
[3 marks]
Question 3 continues on the next page
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Table 2 shows:
• mRNA codons and the amino acid coded for by each codon
• the type of bond formed by the R group of some of the amino acids.
Table 2
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Crystallin is a structural protein found in the human eye. An inherited disease that leads to blindness is caused by changes in properties of crystallin. The replacement of the amino acid Arg with the amino acid Gly causes these changes.
Use information in Table 2 to suggest why this amino acid replacement changes the properties of crystallin.
[2 marks]
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The amino acid replacement of Arg with Gly is caused by a single base substitution mutation in the DNA. The non-mutant DNA triplet is TCC.
Complete Table 3.
Give:
• the mRNA codon complementary to the non-mutant DNA triplet
• the mutated mRNA codon that could cause the change from Arg to Gly in the crystallin protein
• the DNA triplet complementary to this mutated mRNA codon.
[2 marks]
Table 3
mRNA codon for the non-mutant triplet
Mutated mRNA codon
Mutated DNA triplet
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A student dissected an organ from a mammal to observe blood vessels.
He dissected a slice of the organ and identified two blood vessels.
Figure 4 shows a photograph of his dissection.
Figure 4
Figure 5 shows a drawing of the blood vessels from his dissection.
Figure 5
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Suggest two ways the student could improve the quality of his scientific drawing of the blood vessels in this dissection.
[2 marks]
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Identify the type of blood vessel labelled as X and the type of blood vessel labelled as Y in Figure 4.
Describe one feature that allowed you to identify the blood vessels.
[2 marks]
Blood vessel X
Blood vessel Y
Feature
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Describe two precautions the student should take when clearing away after the dissection.
[2 marks]
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Describe how a sample of chloroplasts could be isolated from leaves.
[4 marks]
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Scientists grew two groups of plants:
• control plants with all the inorganic ions needed
• iron-deficient plants with all the inorganic ions needed but without iron ions.
After 1 week, the scientists measured the mass of protein and the mass of chlorophyll in the chloroplasts isolated from samples of leaves of these two groups of plants.
Table 4 shows the scientists’ results.
Table 4
Mass of protein / percentage of control
Mass of chlorophyll / percentage of control
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Some proteins found inside the chloroplast are synthesised inside the chloroplast.
Give one feature of the chloroplast that allows protein to be synthesised inside the chloroplast and describe one difference between this feature in the chloroplast and similar features in the rest of the cell.
[2 marks]
Feature
Structural difference
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The ratio of protein to chlorophyll in control plants is 9:1
Use the information in Table 4 to calculate the ratio of protein to chlorophyll in
iron-deficient plants.
[1 mark]
Ratio
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The scientists also observed the chloroplasts from the samples of leaves using an electron microscope.
Figure 6 shows a chloroplast from a control plant (image A) and a chloroplast from an iron-deficient plant (image B).
Figure 6
Use Figure 6 to suggest why iron-deficient plants have a reduced growth rate.
[3 marks]
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Figure 7 shows the mean distance between centromeres and the poles (ends) of the spindle during mitosis.
Figure 7
Calculate the rate of movement of the centromeres during phase E.
Give your answer in μm minute–1 and to 3 decimal places.
[2 marks]
μm minute–1
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Name the three phases of mitosis shown by C, D and E on Figure 7.
Describe the role of the spindle fibres and the behaviour of the chromosomes during each of these phases.
[5 marks]
C
D
E
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ADCs are molecules made of a monoclonal antibody linked to a cancer drug.
Figure 8 shows how an ADC enters and kills a tumour cell.
The process of entering the cell and the breakdown of the antibody to release the drug is very similar to phagocytosis.
Figure 8
Use your knowledge of phagocytosis to describe how an ADC enters and kills the tumour cell.
[3 marks]
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Some of the antigens found on the surface of tumour cells are also found on the surface of healthy human cells.
Use this information to explain why treatment with an ADC often causes side effects.
[2 marks]
Question 7 continues on the next page
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Scientists investigated whether one type of ADC could be used to treat human breast cancer.
This ADC is a monoclonal antibody combined with a drug to inhibit mitosis. The monoclonal antibody binds to a protein found on human breast cancer cells.
The scientists placed small pieces of human breast cancer tissue under the skin of mice.
The scientists then randomly divided the mice into three groups. They treated the groups as follows on day 0.
Group G – control
Group H – injected with monoclonal antibody only
Group J – injected with ADC (monoclonal antibody + drug).
Every few days, the scientists measured the volume of the tumours formed from the human breast cancer tissue.
Figure 9 shows the scientists’ results.
Figure 9
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Mice in Group H were injected with 2 mg kg–1 of monoclonal antibody.
The monoclonal antibody was in a solution of concentration 500 mg dm–3
Calculate the volume of antibody solution that the scientists would have injected into a 23 g mouse. Give your answer in dm3 and in standard form.
[2 marks]
dm3
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Suggest one reason why there are no data for Group G and Group H after day 8
[1 mark]
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Suggest and explain two further investigations that should be done before this ADC
is tested on human breast cancer patients.
[2 marks]
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2
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Describe how a triglyceride molecule is formed.
[3 marks]
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Table 5 shows some properties of four fatty acids.
Table 5
Fatty acid
Number of carbon atoms in the R group
Number of double bonds in the R group
Caprylic acid
8
0
Palmitoleic acid
16
1
Stearic acid
18
0
Linoleic acid
18
2
Figure 10 shows diagrams of these fatty acids.
Figure 10
Put a tick () in one box that contains correct information about one of these fatty acids.
[1 mark]
Caprylic acid is an unsaturated fatty acid represented by diagram L.
Linoleic acid is a saturated fatty acid represented by diagram N.
Palmitoleic acid is an unsaturated fatty acid represented by diagram K.
Stearic acid is a saturated fatty acid represented by diagram M.
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The percentage of saturated fatty acids compared with unsaturated fatty acids found in lipid stores in seeds differs in different populations.
Scientists investigated two populations of the plant, Helianthus annuus.
The scientists grew young plants from seeds collected from each population. They placed the seeds on wet tissue paper so that the root growth was visible.
They grew seeds from each population at two temperatures:
• warm temperature of 24 °C
• cool temperature of 10 °C
After 10 days, the scientists measured the length of each root.
Table 6 shows some of the properties of the two populations and the scientists’ results.
Table 6
Population
Temperature in natural environment
In the seed – Mean percentage of fatty acids that are saturated
Mean length of root after 10 days at 24 °C / mm
(± 2 x standard deviation)
Mean length of root after 10 days at 10 °C / mm
(± 2 x standard deviation)
1
Warm
10.9
8.2 (±1.0)
3.1 (±0.3)
2
Cool
6.1
5.5 (±0.9)
4.3 (±0.2)
The mean ±2 × standard deviation includes 95% of the data.
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The scientists used a data logger to measure the length of the root rather than a ruler.
Suggest one reason why they used a data logger and explain why this was important in this investigation.
[1 mark]
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It is known that:
• during respiration saturated fatty acids yield more energy than unsaturated fatty acids
• saturated fatty acids have higher melting points than unsaturated fatty acids
• lipases in seeds act more rapidly on liquid substrates.
Use this information and Table 6 to show how each population is better adapted for its natural environment when compared with the other population.
[4 marks]
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Although these two populations are completely separate and show genetic variation, they are both called Helianthus annuus.
Explain why they are both given this name.
[1 mark]
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Complete Table 7 with ticks () to show which elements are found in the following biological molecules.
[2 marks]
Table 7
Biological molecules
Element
Carbon
Nitrogen
Oxygen
Phosphorus
Galactose
Phospholipid
RNA
Sucrose
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After Watson and Crick proposed the model of DNA structure, scientists investigated the possible mechanisms for DNA replication.
Two scientists grew a bacterial population, providing them with a nitrogen source containing only the heavy isotope of nitrogen, 15 N. As soon as all the DNA in this population contained 15 N, the scientists changed the nitrogen source to one containing only the lighter isotope of nitrogen, 14 N. They changed the nitrogen source at 0 hours.
During the investigation, the scientists measured the size of the population of bacterial cells.
Figure 11 shows the scientists’ results.
Figure 11
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The generation time for a population of bacteria is the time taken for all the bacteria to divide once by binary fission.
Use Figure 11 and the following equation to calculate the generation time for this population of bacteria. Give your answer in hours.
[2 marks]
Number of generations= log10size of population at time +4 hourssize of population at time – 4 hourslog10 2
Generation time
hours
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At intervals during this investigation, the scientists removed samples of the bacterial population, isolated the DNA and measured the density of the DNA.
DNA made using 15 N has a higher density than DNA made using 14 N.
Figure 12 shows the scientists’ results.
Figure 12
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There are three possible models of DNA replication.
These models are shown in Figure 13.
Figure 13
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Which of these models, P, Q or R, is supported by the results shown in Figure 12?
Give the letter and name of the model supported and explain why the results do not support the other models.
[3 marks]
Model
Name
Explanation for first unsupported model
Explanation for second
unsupported model
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Describe the structure of DNA.
[5 marks]
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Name and describe five ways substances can move across the cell-surface membrane into a cell.
[5 marks]
Question 10 continues on the next page
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Figure 14 shows transmission electron micrographs of two cells, one animal cell and one prokaryotic cell.
Figure 14
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Contrast the structure of the two cells visible in the electron micrographs shown in Figure 14.
[5 marks]
END OF QUESTIONS
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